# Find the Jordan Canonical Form that is similar with the idempotent matrix A

Find the Jordan Canonical Form that is similar to the idempotent matrix $$A$$.

I know that since $$A=A^2$$ then $$A(A-I)=0$$ so the minimal polynomial is $$m_A(\lambda)=\lambda(\lambda-1)$$.

I also know that the largest Jordan block corresponding to $$\lambda=0$$ and $$\lambda=1$$ have the size $$1 \mathrm x1$$

But I don't really how to continue from here. I would really appreciate some help

• I believe that you have shown that $A$ is diagonalizable with eigenvalues $0$ and $1$. – angryavian Apr 20 at 19:38
• From $A^2=A$ you can deduce that the minimal polynomial divides $\lambda(\lambda-1)$, but you do not know (from just that fact) that it equals this. It could be that the minimal polynomial is $\lambda$ (if $A=0$), or $\lambda-1$ (if $A$ is the identity). – Arturo Magidin Apr 20 at 19:43
• The Jordan canonical form of $A$ is simple. It's a diagonal matrix with only zeros and ones (maybe only ones or only zeros). – amsmath Apr 20 at 19:57

Though cannot conclude about the minimal polynomial as you did, you have already done most of the work. From $$A^2=A$$ you can deduce, as you say, that the only possible eigenvalues are $$0$$ and $$1$$. Next, as you mention, you deduce that the Jordan blocks can only be $$1\times 1$$ (because otherwise they wouldn't be idempotents). So the Jordan form of $$A$$ has $$1\times 1$$ blocks (so it's diagonal) and the diagonal consists of $$0$$ and $$1$$.